Two isolated conducting spheres `S_1 and S_2` of radius `2/3 R and 1/3 R` have `12 mu C` and `-3 muC` charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire . A long time after this is done the charges on `S_1 and S_2` are respectively :

Options

A. `6 mu C and 3 mu C `
B. ` 4.5 mu C ` on both
C. `+4.5 mu C and -4.5 mu C`
D. `1mu C and 6 mu C `

Explanation

To solve the problem, we need to determine the final charges on the two conducting spheres \( S_1 \) and \( S_2 \) after they are connected by a conducting wire. Here are the steps to find the solution:

### Step 1: Calculate the initial total charge
The initial charges on the spheres are:
- Charge on \( S_1 \) ( \( Q_1 \) ) = \( 12 \, \mu C \)
- Charge on \( S_2 \) ( \( Q_2 \) ) = \( -3 \, \mu C \)

The total initial charge \( Q_{initial} \) is given by:
\[ Q_{initial} = Q_1 + Q_2 = 12 \, \mu C + (-3 \, \mu C) = 12 \, \mu C - 3 \, \mu C = 9 \, \mu C \]

### Step 2: Understand the concept of potential
When the two spheres are connected by a conducting wire, charge will flow between them until they reach the same electric potential. The potential \( V \) of a conducting sphere is given by:
\[ V = \frac{k \cdot Q}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

### Step 3: Set up the equations for final charges
Let \( Q_f \) be the final charge on \( S_1 \) and \( Q'_f \) be the final charge on \( S_2 \). Since charge is conserved:
\[ Q_f + Q'_f = Q_{initial} = 9 \, \mu C \]

The radii of the spheres are:
- Radius of \( S_1 \) = \( \frac{2}{3}R \)
- Radius of \( S_2 \) = \( \frac{1}{3}R \)

The potentials of the spheres after they are connected must be equal:
\[ \frac{k \cdot Q_f}{\frac{2}{3}R} = \frac{k \cdot Q'_f}{\frac{1}{3}R} \]

### Step 4: Simplify the potential equation
Canceling \( k \) and \( R \) from both sides gives:
\[ \frac{Q_f}{\frac{2}{3}} = \frac{Q'_f}{\frac{1}{3}} \]
This simplifies to:
\[ 3Q_f = 2Q'_f \]

### Step 5: Solve the equations
Now we have two equations:
1. \( Q_f + Q'_f = 9 \)
2. \( 3Q_f = 2Q'_f \)

From the second equation, we can express \( Q'_f \) in terms of \( Q_f \):
\[ Q'_f = \frac{3}{2}Q_f \]

Substituting this into the first equation:
\[ Q_f + \frac{3}{2}Q_f = 9 \]
\[ \frac{5}{2}Q_f = 9 \]
\[ Q_f = \frac{9 \times 2}{5} = \frac{18}{5} \, \mu C = 3.6 \, \mu C \]

Now substituting back to find \( Q'_f \):
\[ Q'_f = 9 - Q_f = 9 - 3.6 = 5.4 \, \mu C \]

### Final Charges
Thus, the final charges on the spheres are:
- Charge on \( S_1 \) ( \( Q_f \) ) = \( 3.6 \, \mu C \)
- Charge on \( S_2 \) ( \( Q'_f \) ) = \( 5.4 \, \mu C \)